MCAT Organic Chemistry Review

Spectroscopy

11.1 Infrared Spectroscopy

Infrared (IRspectroscopy measures molecular vibrations, which can be seen as bond stretching, bending, or combinations of different vibrational modes. To record an IR spectrum, infrared light is passed through a sample, and the absorbance is measured. By determining what bonds exist within a molecule, we hope to infer the functional groups in the molecule.

INTRAMOLECULAR VIBRATIONS AND ROTATIONS

The infrared light range runs from λ = 700 nm to 1 mm, but the useful absorptions for spectroscopy occur at wavelengths of 2500 to 25,000 nm. On an IR spectrum, we use an analog of frequency called wavenumber. The standard range corresponding to 2500 to 25,000 nm is 4000 to 400 cm–1. When light of these wavenumbers is absorbed, the molecules enter excited vibrational states. Four types of vibration that can occur are shown in Figure 11.1. Others include twisting and folding.

Figure 11.1. Molecular Vibrations Measured by Infrared (IR) Spectroscopy Bond bending and stretching; twisting and folding can also occur.

KEY CONCEPT

Wavenumbers (cm–1) are an analog of frequency.  whereas 

More complex vibration patterns, caused by the motion of the molecule as a whole, can be seen in the 1500 to 400 cm–1 range. This is called the fingerprint region because the specific absorbance pattern is characteristic of each individual molecule. Spectroscopy experts can use this region to identify a substance, but you won’t ever need to use it on the MCAT.

For an absorption to be recorded, the vibration must result in a change in the bond dipole moment. This means that molecules that do not experience a change in dipole moment, such as those composed of atoms with the same electronegativity or molecules that are symmetrical, do not exhibit absorption. For example, we cannot get an absorption from O2 or Br2, but we can from HCl or CO. Symmetric bonds, such as the triple bond in acetylene (C2H2), will also be silent.

KEY CONCEPT

Symmetric stretches do not show up in IR spectra because they involve no net change in dipole movement.

CHARACTERISTIC ABSORPTIONS

For the MCAT, you only need to memorize a few absorptions. The first is the hydroxyl group, O–H, which absorbs around 3300 cm–1 with a broad (wide) peak. The second is the carbonyl, which absorbs around 1700 cm–1 with a sharp (deep) peak. In Table 11.1, notice how the bond between any atom and hydrogen always has a relatively high absorption frequency and how, as we add more bonds between carbon atoms, the absorption frequency increases. N–H bonds are in the same region as O–H bonds (around 3300 cm–1), but have a sharp peak instead of a broad one. You should be able to identify these three peaks in an IR spectrum. If you need to identify other peaks on Test Day, a list or table of peak wavenumbers will be provided.

Functional Group

Wavenumber (cm−1)

Vibration

Alkanes

2800–3000
1200

C–H
C–C

Alkanes

3080–3140
1645

=C–H
C=C

Alkynes

3300
2200

≡C–H
C≡C

Aromatic

2900–3100
1475–1625

C–H
C–C

Alcohols

3100–3500

O–H(broad)

Ethers

1050–1150

C–O

Aldehydes

2700–2900
1700–1750

(O)C–H
C=O

Ketones

1700–1750

C=O

Carboxylic Acids

1700–1750
2800–3200

C=O
O–H (broad)

Amines

3100–3500

N–H (sharp)

Table 11.1. Absorption Frequencies

MCAT EXPERTISE

Infrared spectroscopy is best used for identification of functional groups. The most important peaks to know are:

·        O–H (broad around 3300 cm–1)

·        N–H (sharp around 3300 cm–1)

·        C=O (sharp around 1750 cm–1)

We can learn a great deal of information from an IR spectrum; for the MCAT, all of the information comes from the frequencies between 1400 and 4000 cm–1. Everything lower (in the fingerprint region) is out of scope. IR spectra are plotted as percent transmittance, the amount of light that passes through the sample and reaches the detector, vs. wavenumber.

Figure 11.2. IR Spectrum of an Aliphatic Alcohol Broad peak at 3300 cm–1: –OH.

KEY CONCEPT

In an IR spectrum, percent transmittance is plotted vs. frequency. The equation relating absorbance, A, and percent transmittance, % T, is A = 2 –log % T; this means that maximum absorptions appear as the bottom of valleys on the spectrum.

Figure 11.2 shows the IR spectrum for an aliphatic alcohol: the large broad peak at 3300 cm–1 is due to the presence of a hydroxyl group, whereas the sharper peak at 3000 cm–1 is due to the carbon–hydrogen bonds in the alkane portion of the molecule, as listed in Table 11.1.

MCAT Concept Check 11.1:

Before you move on, assess your understanding of the material with these questions.

1.    What does infrared (IR) spectroscopy measure? What is IR spectroscopy generally used for?

2.    What two peaks would you expect to see in the IR spectrum of a carboxylic acid?